A=1+2+4+8+...+4098
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\(B=1+2+4+...+4096\)
\(B=1+2+2^2+2^3+...+2^{12}\)
\(2B=2+2^2+...+2^{13}\)
\(2B-B=\left(2+2^2+...+2^{13}\right)-\left(1+2+...+2^{12}\right)\)
\(B=2^{13}-1\)
B=1+2+4+8+...+4098
B=\(1+2+2^2+2^3+....+2^{12}\)
2B=\(2+2^2+2^3+2^4+.....+2^{13}\)
2B-B=\(\left(2+2^2+2^3+2^4+...+2^{13}\right)-\left(1+2+2^2+2^3...+2^{12}\right)\)
B=\(2^{13}-1\)
a; 100 - 94 + 90 - 84 + 80 - 74 + 70 - 64 + 60 - 54
= (100 - 94) + (90 - 84) + (80 - 74) + (70 - 64) + (60 - 54)
= 6 + 6 + 6 + 6 + 6
A = 6 x 5
A = 30
b; 5.125.20.16
= (5.20). (125.8).2
= 100.1000.2
= 200000
\(4\dfrac{3}{8}+5\dfrac{2}{3}=\dfrac{4\times8+3}{8}+\dfrac{5\times3+2}{3}\\ =\dfrac{35}{8}+\dfrac{17}{3}=\dfrac{35\times3+17\times8}{24}=\dfrac{241}{24}\\ ----\\ 2\dfrac{3}{8}+1\dfrac{1}{4}=\dfrac{2\times8+3}{8}+\dfrac{1\times4+1}{4}=\dfrac{19}{8}+\dfrac{5}{4}\\ =\dfrac{19+5\times2}{8}=\dfrac{29}{8}\\ ----\\ 2\dfrac{3}{8}-1\dfrac{1}{4}+5\dfrac{1}{3}=\left(2-1+5\right)+\left(\dfrac{3}{8}-\dfrac{1}{4}+\dfrac{1}{3}\right)\\ =6+\dfrac{3\times3-1\times6+1\times8}{24}\\ =6+\dfrac{11}{24}=6\dfrac{11}{24}\)